Question

in some cases, there may be a mixture of two or more acids. When this happens, the stronger acid will dissociate completely, so you should be able to calculate how much H,0 is in solution as a result. Writing out the equation for the dissociation often helps The weak acid will attempt to dissociate and is governed by the equilibrium (K,). You should write out the equation for the weak acid Now-write in the values for the concentrations that you'll use for the weak acid ICE table. of H,0 is not zero. It will be the concentration thot wos produced from the strong ocid. Be careful, your starting concentration What is the pH (and [H,O*1 ) for a 0.100 M formic acid and 0.05M HBr solution (Formic acid K= 1.8 x 103 value, check the relative strengths of these acids) Determine a conjugate acid and the conjugate base for each of the following compounds: AY Conjugate acid Conjugate base нсо- HРО, HS E, S H2O If the following equilibria have known K values, write the equations that would have a corresponding Ka NH3H20 NH+OH HS+H20 H2S+OH CLO- H20 HCLO OH H,PO, write the equations that correspond to Ka Kaz Ka For the polyprotic acid: 11" Now if you imagine the conjugate bases acting as a base, HPO,H2PO, PO Write out the equations of each of these conjugate bases acting as a base (react with water to produce a conjugate acid and hydronium. Which of the bases would have the first proton accepted on the way to becoming fully protonated again? Which of the bases would have the last proton to be accepted before returning back to H3PO4? These are defined as K, K, K corresponds to which Kp? as the first proton to be accepted, the second, and third, etc-which of the equations Try to match which Ka is related to which Kb

Answer 1

1) You have the expression of Ka:

Ka = [H3O +] * [A-] / [HA] = (0.05 + X) * X / 0.1 = 1.8x10 ^ -5

It clears X = 3.6x10 ^ -5 M

The total concentration and pH are calculated:

[H3O +] = 0.05 + 3.6x10 ^ -5 = 0.050036 M

pH = - log 0.050036 = 1.3

2) The conjugate bases are:

i) CO3-2

ii) HPO4-2

iii) S-2

iv) OH-

3) The expressions are written:

i) NH4 + + H2O = NH3 + H3O +

ii) H2S + H2O = HS- + H3O +

iii) HClO + H2O = ClO- + H3O +

4) The equations are written:

Ka1 = [H2PO4-] * [H +] / [H3PO4]

Ka2 = [HPO4-2] * [H +] / [H2PO4-]

Ka3 = [PO4-3] * [H +] / [HPO4-2]

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